Django URL TypeError:在include()的情况下,view必须是可调用的或list/tuple

Question

升级到Django 1.10后,我收到错误:

TypeError: view must be a callable or a list/tuple in the case of include().

我的urls.py如下:

from django.conf.urls import include, url

urlpatterns = [
    url(r'^$', 'myapp.views.home'),
    url(r'^contact/$', 'myapp.views.contact'),
    url(r'^login/$', 'django.contrib.auth.views.login'),
]

完整的追溯是:

Traceback (most recent call last):
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/autoreload.py", line 226, in wrapper
    fn(*args, **kwargs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/commands/runserver.py", line 121, in inner_run
    self.check(display_num_errors=True)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py", line 385, in check
    include_deployment_checks=include_deployment_checks,
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py", line 372, in _run_checks
    return checks.run_checks(**kwargs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/registry.py", line 81, in run_checks
    new_errors = check(app_configs=app_configs)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py", line 14, in check_url_config
    return check_resolver(resolver)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py", line 24, in check_resolver
    for pattern in resolver.url_patterns:
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py", line 35, in __get__
    res = instance.__dict__[self.name] = self.func(instance)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py", line 310, in url_patterns
    patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py", line 35, in __get__
    res = instance.__dict__[self.name] = self.func(instance)
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py", line 303, in urlconf_module
    return import_module(self.urlconf_name)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py", line 37, in import_module
    __import__(name)
  File "/Users/alasdair/dev/urlproject/urlproject/urls.py", line 28, in <module>
    url(r'^$', 'myapp.views.home'),
  File "/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/conf/urls/__init__.py", line 85, in url
    raise TypeError('view must be a callable or a list/tuple in the case of include().')
TypeError: view must be a callable or a list/tuple in the case of include().

Answer 1

Django 1.10不再允许您'myapp.views.home'在URL模式中将视图指定为字符串(例如).

解决方案是更新您urls.py的视图可调用.这意味着您必须导入您的视图urls.py.如果您的URL模式没有名称,那么现在是添加名称的好时机,因为使用虚线python路径进行反转不再有效.

from django.conf.urls import include, url

from django.contrib.auth.views import login
from myapp.views import home, contact

urlpatterns = [
    url(r'^$', home, name='home'),
    url(r'^contact/$', contact, name='contact'),
    url(r'^login/$', login, name='login'),
]

如果有很多视图,那么单独导入它们可能会很不方便.另一种方法是从您的应用程序导入视图模块.

from django.conf.urls import include, url

from django.contrib.auth import views as auth_views
from myapp import views as myapp_views

urlpatterns = [
    url(r'^$', myapp_views.home, name='home'),
    url(r'^contact/$', myapp_views.contact, name='contact'),
    url(r'^login/$', auth_views.login, name='login'),
]

请注意,我们使用了as myapp_views和as auth_views,它允许我们views.py从多个应用程序导入而不会发生冲突.

有关的更多信息,请参阅Django URL调度程序文档urlpatterns.